What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)2(aq) solution

32.29 mL

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2HCl(aq)+ Ba(OH)2(aq) —> BaCl2(aq) + 2H2O

From the balanced equation above, the following were obtained:

Mole ratio of the acid (nA) = 2

Mole ratio of the base (nB) = 1

Next, data were obtained from. This include the following:

Molarity of acid, HCl (Ma) = 0.3428 M

Volume of acid, HCl (Va) =.?

Molarity of base, Ba(OH)2 (Mb) = 0.2350 M

Volume of base, Ba(OH)2 (Vb) = 23.55 mL

Finally, we shall determine the volume of the acid, HCl required for the reaction as follow:

MaVa /MbVb = nA/nB

0.3428 × Va / 0.2350 × 23.55 = 2/1

Cross multiply

0.3428 × Va = 0.2350 × 23.55 × 2

Divide both side by 0.3428

Va = (0.2350 × 23.55 × 2) / 0.3428

Va = 32.29 mL

Therefore, the volume of acid, HCl required for the reaction is 32.29 mL

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